(2)因为0。ABC=。0CSnH=5V5,的得bC=12V3,2由余弦定理得a2=分+e2-2bco0sA,(25=+c-2x125x5,故+c=48,b+e=6+c2+2c=48+245-24+25-=20+5-6+25,图正弦定理可得a一=6。=s。=4√3,所以snB=bsinA sin B=45mc=45sin B+sinc=b+c=6+23+14W34√3216.【解析】(1)因为PA⊥平面ABCD,而ADC平面ABCD,所以PA⊥AD,又AD⊥PB,PB∩PA=P,PB,PAC平面PAB,